Problem: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $24$ units long $\overline{AB}$ is $26$ units long What is $\sec(\angle BAC)?$ $A$ $C$ $B$ $10$ $24$ $26$
Explanation: $\sec(\angle BAC) = \dfrac{1}{\cos(\angle BAC)}$ How can we find $\cos(\angle BAC)$ SOH CAH TOA osine = djacent over ypotenuse Adjacent $= \overline{AC} = 10$ Hypotenuse $= \overline{AB} = 26$ $\cos(\angle BAC) = \dfrac{10}{26}$ $\sec(\angle BAC) = \dfrac{1}{\cos(\angle BAC)} = \dfrac{26}{10}$